# What's the probability that Abe will win the dice game?

Abe and Bill are playing a game. A die is rolled each turn.

If the die lands 1 or 2, then Abe wins.

If the die lands 3, 4, or 5, then Bill wins.

If the die lands 6, another turn occurs.

What's the probability that Abe will win the game?

I think that the probability is $\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.

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As a visitor from rpg.stackexchange.com, I can say that it's very common to generate dice which you don't have (such as a d3) by saying "Roll a d4 and reroll 4's". Effectively, any result you reroll from scratch doesn't exist on the die. I wouldn't have been able to state it formally, though, which is why I'm just a visitor here! – Bobson Oct 4 '12 at 4:25
@Bobson: For a d3, wouldn't you normally take a d6 and map 1-2 to 1, 3-4 to 2, and 5-6 to 3? – Joren Oct 4 '12 at 14:03
@Joren - I've done it both ways. The problem with the d6 is sometimes you forget to declare whether you're mapping it that way, or whether you're mapping it 4->1, 5->2, 6->3. Which is also perfectly valid. With the d4, there's no question about it. Admittedly, it may not have been the best choice of examples. – Bobson Oct 4 '12 at 14:11
Could someone with enough rep change this to say die instead of dice. It wont let me save the suggested edits because there aren't enough characters changed. – zzzzBov Oct 4 '12 at 15:26

You are right. You can just ignore rolls of $6$ as they leave you back in the same situation. To formalize this, the chance Abe wins on turn $n$ is $\frac 13 \left(\frac 16 \right)^{n-1}$ and the chance that Bill wins on turn $n$ is $\frac 12 \left(\frac 16 \right)^{n-1}$. You can sum these if you want.

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Let $P$ be the chance that Abe wins, then we have

$$P=\frac{1}{3}+\frac{1}{6}P$$

Solving P from this equation we get

$$P=\frac{2}{5}$$

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I agree with you that $\dfrac{2}{5}$ is obvious. End of story. But if you really want to sum a series, abbreviate by $A$ the event "$1$ or $2$" and by $S$ the event "$6$." Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on.

These have probabilities $\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{2}{6}$, and so on. So we want to sum the series $$a+ar+ar^2+ar^3+\cdots,$$ where $a=\dfrac{2}{6}$ and $r=\dfrac{1}{6}$.

By the usual formula for the sum of an infinite geometric series, this is $\dfrac{a}{1-r}$, which simplifies to $\dfrac{2}{5}$.

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The key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before. So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$p_A = \frac 26 + \frac 16 p_A\\ p_B = \frac 36 + \frac 16 p_B$$ which is easy to solve.

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If you want to use geometric series, you can do the following.

$$\text{P(Abe wins) = P(Abe wins in the first roll) + P(Abe wins in the second roll) + } \dots$$ $$=\frac{2}{6}+\frac{1}{6}\frac{2}{6} + \left(\frac{1}{6}\right)^2 \frac{2}{6} + \dots$$ $$=\frac{1}{3}\left(1+\frac{1}{6}+\left(\frac{1}{6}\right)^2 + \dots\right)$$ $$=\frac{1}{3}\frac{1}{1-\frac{1}{6}}$$ $$=\frac{2}{5}$$

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The chance Abe win is 2/5

You can forget about the reroll. At each reroll, the chance that Abe win is 2/5. So no matter how many rerolls out there, the chance that Abe will win is the weighted average of chance that Abe will win multiplied by probability distribution of the rerolls number.

That average is easy to compute because it's always 2/5.

All answers are the same but I try to do this as intuitive as possible because I can't type equation :)

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